# The Unapologetic Mathematician

## Permutation Representations from Partitions

Now that we have an action of $S_n$ on the Young tabloids of shape $\lambda\vdash n$, we can consider the permutation representation $M^\lambda$ that corresponds to it. Let’s consider a few examples.

First, let $\lambda=(n)$. This is a pretty trivial “partition”, consisting of one piece of length $n$. The Ferrers diagram of $\lambda$ looks like

$\displaystyle\begin{array}{cccc}\bullet&\bullet&\cdots&\bullet\end{array}$

Any Young tableau thus contains all $n$ numbers on the single row, so they’re all row-equivalent. There is only one Young tabloid:

$\displaystyle\begin{array}{cccc}\cline{1-4}1&2&\cdots&n\\\cline{1-4}\end{array}$

We conclude that $M^{(n)}$ is a one-dimensional vector space with the trivial action of $S_n$.

Next, let $\lambda=(1^n)$ — another simple partition with $n$ parts of length $1$ each. The Ferrers diagram looks like

$\displaystyle\begin{array}{c}\bullet\\\bullet\\\vdots\\\bullet\end{array}$

Now in every Young tableau each number is on a different line, so no two tableaux are row-equivalent. They each give rise to their own Young tabloid, such as

$\displaystyle\begin{array}{c}\cline{1-1}1\\\cline{1-1}2\\\cline{1-1}\vdots\\\cline{1-1}n\\\cline{1-1}\end{array}$

These tabloids correspond to permutations; a generic one looks like

$\displaystyle\begin{array}{c}\cline{1-1}\pi(1)\\\cline{1-1}\pi(2)\\\cline{1-1}\vdots\\\cline{1-1}\pi(n)\\\cline{1-1}\end{array}$

The action of $S_n$ on these tabloids is basically the same as left-multiplication on the underlying set $S_n$. And so we find the left regular representation.

Finally, consider the partition $\lambda=(n-1,1)$. This time the Ferrers diagram looks like

$\displaystyle\begin{array}{cccc}\bullet&\bullet&\cdots&\bullet\\\bullet&&&\end{array}$

and a sample Young tabloid looks like

$\displaystyle\begin{array}{cccc}\cline{1-4}1&2&\cdots&n-1\\\cline{1-4}n&&&\\\cline{1-1}\end{array}$

Any Young tabloid of shape $\lambda$ is uniquely specified by the single entry on the second row. Any permutation shuffles them around exactly like it does these entries, and so $M^{(n-1,1)}$ is isomorphic to the defining representation.

December 14, 2010 -

1. [...] try to calculate the characters of the Young tabloid modules we’ve constructed. Since these come from actions of on various sets, we have our usual [...]

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2. [...] we’re interested in. For any partition , the Specht module is the submodule of the Young tabloid module spanned by the polytabloids where runs over the Young tableaux of shape [...]

Pingback by Specht Modules « The Unapologetic Mathematician | December 27, 2010 | Reply

3. [...] space with the trivial group action. This is the only possibility anyway, since , and we’ve seen that is itself a one-dimensional vector space with the trivial action of [...]

Pingback by Examples of Specht Modules « The Unapologetic Mathematician | December 28, 2010 | Reply

4. [...] be a submodule of one of the Young tabloid modules. Then I say that either contains the Specht module , or it is contained in the orthogonal [...]

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5. [...] in the case we care about the space is the Young tabloid module , with the basis of Young tabloids having the dominance ordering. In particular, we consider for [...]

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6. [...] defined the Specht module as the subspace of the Young tabloid module spanned by polytabloids of shape . But these polytabloids are not independent. We’ve seen [...]

Pingback by Standard Polytabloids Span Specht Modules « The Unapologetic Mathematician | January 21, 2011 | Reply

7. [...] of generalized tableaux of the vector space is in bijection with the basis of -tabloids of the vector space . And this space carries an action of — the linear extension of the action on tabloids. We [...]

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